91Ó°ÊÓ

A particle is executing a simple harmonic motion.

a continuous random variable,whose integral across an interval offers the likelihood that the variable's value falls inside the same interval.

Function x is given as .

$\begin{array}{c}v=\frac{dx}{dt}\\ =−a\mathrm{Ó\neg }\sin \left(\mathrm{Ó\neg }t\right)\\ =−a\mathrm{Ó\neg }\sqrt{1−{\cos }^2\left(\mathrm{Ó\neg }t\right)}\end{array}$

The value of $\frac{dx}{dt}$is given below.

$\begin{array}{c}\frac{dx}{dt}=−a\mathrm{Ó\neg }\sqrt{1−{\left(\frac{x}{a}\right)}^2}\\ =−\mathrm{Ó\neg }\sqrt{a^2−x^2}\end{array}$

$\begin{array}{c}dt=\frac{dx}{v}\\ =−\frac{dx}{\mathrm{Ó\neg }\sqrt{a^2−x^2}}\end{array}$

The probability of function is given below.

$\begin{array}{c}f\left(x\right)dx=\frac{dx}{\sqrt{a^2−x^2}}\\ f\left(x\right)=\frac{C}{\sqrt{a^2−x^2}}\end{array}$

Find the value of constant.

$\begin{array}{c}{∫}_{−a}^a\frac{Cdx}{\sqrt{a^2−x^2}}=C{\left[\arccos \left(\frac{x}{a}\right)\right]}_{−a}^a\\ =C\mathrm{π}\\ C\mathrm{π}=1\\ C=\frac{1}{\mathrm{π}}\end{array}$

Substitute the value of C the function becomes as follows.

$f\left(x\right)=\frac{1}{\mathrm{π}\sqrt{a^2−x^2}}$

The graphs are shown below.

The average of the function is mentioned below

$\begin{array}{c}\mathrm{μ}={∫}_{−\mathrm{∞}}^{\mathrm{∞}}xf\left(x\right)dx\\ ={∫}_{−\mathrm{∞}}^{\mathrm{∞}}\frac{xdx}{\mathrm{π}\sqrt{a^2−x^2}}\\ ={\left[−\frac{\sqrt{a^2−x^2}}{\mathrm{π}}\right]}_{−\mathrm{∞}}^{\mathrm{∞}}\\ =0\end{array}$

The variance is given below.

$\begin{array}{c}\mathrm{Var}\left(x\right)={∫}_{−\mathrm{∞}}^{\mathrm{∞}}{x}^{2}f\left(x\right)dx\\ ={∫}_{−a}^a\frac{x^{2}dx}{\mathrm{π}\sqrt{a^2−x^2}}\\ ={\left[\frac{a^2\arcsin \left(\frac{x}{\left|a\right|}\right)−x\sqrt{a^2−x^2}}{2\mathrm{π}}\right]}_{−a}^a\\ =\frac{a^2}{2}\end{array}$

The standard deviation is given below.

$\begin{array}{c}\mathrm{σ}=\sqrt{\mathrm{var}\left(x\right)}\\ =\sqrt{\frac{a^2}{2}}\\ =\frac{a}{\sqrt{2}}\end{array}$

The required values are mentioned below.

$\begin{array}{c}f\left(x\right)=\frac{1}{\mathrm{π}\sqrt{a^2−x^2}}\\ \mathrm{μ}=0\\ \mathrm{Var}\left(\mathrm{x}\right){}_{-\mathrm{a}}^{\mathrm{a}}=\frac{a^2}{2}\\ \mathrm{σ}=\frac{a}{\sqrt{2}}\end{array}$